“一些”恒等式

其实到目前就写了俩……见到的话可能还会更新吧，不过马上就退役了，大概也见不到了

婆罗摩笈多－斐波那契恒等式

[egin{aligned}(a^2+b^2)(c^2+d^2)&=(ac-bd)^2+(ad+bc)^2\&=(ac+bd)^2+(ad-bc)^2end{aligned} ]

证明

将式子展开即可。

[egin{aligned}(a^2+b^2)(c^2+d^2)&=a^2c^2+a^2d^2+b^2c^2+b^2d^2+2a^2b^2c^2d^2-2a^2b^2c^2d^2\&=(a^2c^2+b^2d^2+2a^2b^2c^2d^2)+(a^2d^2+b^2c^2-2a^2b^2c^2d^2)\&=(ac+bd)^2+(ad-bc)^2\&=(a^2c^2+b^2d^2-2a^2b^2c^2d^2)+(a^2d^2+b^2c^2+2a^2b^2c^2d^2)\&=(ac-bd)^2+(ad+bc)^2end{aligned} ]

拉格朗日恒等式

[(sumlimits_{i=1}^na_i^2)(sumlimits_{i=1}^nb_i^2)=(sumlimits_{i=1}^na_ib_i)^2+sumlimits_{1le i<jle n}(a_ib_j-a_jb_i)^2 ]

证明

用非常复杂的方法证明了这东西……

首先：

1. (sumlimits_{1le i<jle n}x=sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^nx)

   容易理解……

2. ((sumlimits_{i=1}^na_ib_i)^2=sumlimits_{i=1}^na_i^2b_i^2+2sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_ib_ia_jb_j)

   表示多项的平方和为每项的平方加上任意两项之间的的乘积的 (2) 倍的和

然后化简式子：

[egin{aligned}&(sumlimits_{i=1}^na_i^2)(sumlimits_{i=1}^nb_i^2)-(sumlimits_{i=1}^na_ib_i)^2\=&sumlimits_{i=1}^nsumlimits_{j=1}^{n}a_i^2b_{j}^2-(sumlimits_{i=1}^na_ib_i)^2\=&(sumlimits_{i=1}^na_i^2b_i^2+sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_i^2b_j^2+sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_j^2b_i^2)-(sumlimits_{i=1}^na_i^2b_i^2+2sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_ib_ia_jb_j)\=&sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_i^2b_j^2+sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_j^2b_i^2-2sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^na_ib_ia_jb_j\=&sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^n(a_i^2b_j^2+a_j^2b_i^2-2a_ia_jb_ib_j)\=&sumlimits_{i=1}^{n-1}sumlimits_{j=i+1}^n(a_ib_j-a_jb_i)^2\=&sumlimits_{1le i<jle n}(a_ib_j-a_jb_i)^2end{aligned} ]

证毕。

用拉格朗日恒等式证明柯西不等式

柯西不等式的形式如下：

[(sumlimits_{i=1}^na_i^2)(sumlimits_{i=1}^nb_i^2)geq (sumlimits_{i=1}^na_ib_i)^2 ]

现在我们已知拉格朗日恒等式成立，即

[(sumlimits_{i=1}^na_i^2)(sumlimits_{i=1}^nb_i^2)=(sumlimits_{i=1}^na_ib_i)^2+sumlimits_{1le i<jle n}(a_ib_j-a_jb_i)^2 ]

又因为 (sumlimits_{1le i<jle n}(a_ib_j-a_jb_i)^2) 的值 (ge0)，所以有

[(sumlimits_{i=1}^na_i^2)(sumlimits_{i=1}^nb_i^2)geq (sumlimits_{i=1}^na_ib_i)^2 ]

即柯西不等式。